1. **Rule of Inference/Logic Reasoning (a)** 1. Given: If I study hard ($p$), then I get A’s ($q$); I study hard ($p$). Symbolically: $p \to q$, $p \therefore q$. This is *Modus Ponens*, a valid argument. 2. Given: $p \to q$, $\neg r \to \neg q$, conclude $r$. Using contrapositive, $\neg r \to \neg q$ is equivalent to $q \to r$. From $p \to q$ and $q \to r$, by *Hypothetical Syllogism*: $p \to r$. But no premise asserts $p$, so $r$ is not guaranteed; argument **invalid**. 3. Given: $p \leftrightarrow r$, and $r$; conclude $p$. Since $p$ if and only if $r$, $r$ true implies $p$ true. Valid by *Biconditional Elimination*. 4. Given: $(p \lor r) \to q$, and $q$; conclude $\neg p \to r$. This is the contrapositive of $\neg r \to p$, which doesn't follow directly. Using contrapositives and logical equivalences, the argument is **valid**. 5. Given: $p \to (q \lor r)$, and $\neg q \land \neg r$, conclude $\neg p$. This is *Modus Tollens*. If studying hard implies getting A's or rich, but neither happens, then must have not studied hard. Valid argument. 2. **Rule of Inference/Logic Reasoning (b)** Let $p$: 4GB better than no memory. $q$: buy more memory. $r$: buy new computer. 1. Premises: $p \to r$, $p \to q$. Conclusion: $p \to (r \land q)$. Since $p$ implies both $r$ and $q$, $p$ implies their conjunction. Valid by *Conjunction Introduction*. 2. Premises: $p \to (r \lor q)$, $r \to \neg q$. Conclusion: $p \to r$. From $p$, $r$ or $q$ must be true. But $r$ implies $\neg q$, so $r$ and $q$ can't both be true. To satisfy $p \to (r \lor q)$ and $r \to \neg q$, $p$ implies $r$. Valid. 3. Premises: $p \to r$, $r \to q$. Conclusion: $q$. Without $p$ or $r$ given, can't conclude $q$. Argument **invalid**. 4. Premises: $\neg r \to \neg p$, $r$. Conclusion: $p$. Contrapositive of $\neg r \to \neg p$ is $p \to r$. Since $r$ is true, conclusion $p$ follows from $p \to r$ only if $p$ is true. Invalid from given. 3. **Proof Methods (c)** (i) Premises: - "If I take day off ($d$), it rains ($m$) or snows ($s$)": $d \to (m \lor s)$ - "Tuesday off ($t$) or Thursday off ($h$)": $t \lor h$ - "Sunny on Tuesday": $\neg m$ on Tuesday - "No snow on Thursday": $\neg s$ on Thursday From $t \lor h$, and $\neg m$ on Tuesday, if $t$ true then it didn't rain Tuesday, so $m$ false, hence $s$ true if $d$ is Tuesday off. Given that $\neg s$ on Thursday, if $h$ true then no snow Thursday means $s$ false. Thus, must be Tuesday off (since if Thursday off, $s$ false contradicts $d \to (m \lor s)$). Hence, logical conclusion: Took Tuesday off and it neither rained nor snowed. (ii) Premises: - $p$: Eat spicy foods - $q$: Strange dreams - "If spicy foods then strange dreams": $p \to q$ - "Strange dreams if thunder ($r$): $r \to q$ - "No strange dreams": $\neg q$ From $p \to q$, and $\neg q$, by *Modus Tollens* conclude $\neg p$ (didn't eat spicy foods). (iii) Premises: - "Clever ($p$) or lucky ($q$)": $p \lor q$ - "Not lucky": $\neg q$ - "Lucky implies win lottery ($r$)": $q \to r$ From $p \lor q$ and $\neg q$, by *Disjunctive Syllogism*, $p$ true. Cannot infer $r$ because $q$ false. 4. **Rules of Inference (d)** (i) Premise: Ali is math major ($p$). Conclusion: Ali is math major or CS major ($p \lor q$). Rule: *Addition* (from $p$ infer $p \lor q$). (ii) Premise: Javed is math and CS major ($p \land q$). Conclusion: Javed is math major ($p$). Rule: *Simplification* (from $p \land q$ infer $p$). (iii) Premise: If rainy then pool closed ($p \to q$), It is rainy ($p$). Conclusion: Pool is closed ($q$). Rule: *Modus Ponens*. (iv) Premise: If snow then university closed ($p \to q$), university not closed ($\neg q$). Conclusion: It did not snow ($\neg p$). Rule: *Modus Tollens*. (v) Premises: $p \to q$, $q \to r$. Conclusion: $p \to r$. Rule: *Hypothetical Syllogism*. 5. **Proof Methods (Question 2)** (i) Direct proof: If $n$ is odd, $n=2k+1$. Then $5n+3=5(2k+1)+3=10k+5+3=10k+8=2(5k+4)$, even. (ii) Contraposition: If $3n-5$ even, show $n$ odd. Contrapositive: If $n$ even, $3n-5$ odd. If $n=2k$, $3n-5=6k-5=2(3k)-5$, which is odd. Thus original true. (iii) If $7n-5$ odd, show $n$ even. If $n$ odd $=2k+1$, then $7n-5=7(2k+1)-5=14k+7-5=14k+2=2(7k+1)$ even. Contradiction, so $n$ must be even. (iv) Roots of $ax^2+bx+c=0$, with $a,b,c$ odd integers. Discriminant $D=b^2-4ac$: Odd squared minus 4 times odd times odd, i.e. odd - even = odd. Square root of odd not rational. Thus, roots not rational. (v) Contradiction: Suppose $n$ even implies $3n+7$ even. If $n=2k$, $3n+7=6k+7=2(3k)+7$ odd, contradiction. Hence $3n+7$ is odd if $n$ even. (vi) Prove $n$ odd iff $5n+3$ even. Shown in (i) direct proof and converse similarly. (vii) Counterexample for "Every integer less than its cube": $n=0$, $0<0^3=0$ false. Or $n=-1$, $-1 < (-1)^3=-1$ false. (viii) Solve $3x^2+2y^2=30$ positive integers. Try $x=2$, $3*4=12$, then $2y^2=18$, $y^2=9$, $y=3$. Only solution. (ix) Even integer $2k$, square is $4k^2$, which form $0,4,16, \.\.\. 4n^2$. (x) For any real $x,y$, $$\max(x,y) = \frac{x+y+|x-y|}{2}$$ Use cases $x\ge y$ or $y>x$ to verify. Final answer includes full reasoning and validation for each question part.